# A Theorem of Dimension of Fiber in Qing Liu (Theorem 4.3.12)

I'm reading Qing Liu. I feel confused about the proof of theorem 4.3.12 and corollary 3.14.

Theorem 4.3.12. Let $$f$$ be a morphism of locally Noetherian schemes. Let $$x\in X$$ and $$y= f(x)$$. Then $$\dim (\mathcal{O}_{X_y,x})\geq \dim (\mathcal{O}_{X,x})-\dim (\mathcal{O}_{Y,y})$$. If, moreover, $$f$$ is flat, then we have equality.

My question:

1.(red sentences) How do we know $$t$$ is not invertible in $$B=\mathcal O_{X,x}$$?

2.(green sentence) After base change with respect to $$Y'$$, $$X'\rightarrow X$$ is a closed immersion. But why $$x$$ is in $$X' = X\times_YY'$$?

Corollary 4.3.14. Let $$f:X\rightarrow Y$$ be a flat subjective morphism of algebraic varieties. We suppose that $$Y$$ is irreducible and that $$X$$ is equidimensional. Then for every $$y\in Y$$, the fiber $$X_y$$ is equidimensional, and we have $$\dim(X_y) = \dim(X)-\dim(Y)$$

My Question: Qing Liu defines algebraic variety as Noetherian scheme of finite type over $$k$$. That means $$X_i$$ may not be reduced. But Proposition 2.5.23(a) only works for domains. In this case how to get the highlight equality?

Theorem 2.5.15 Let $$(A,\frak m)$$ be a Noetherian local ring, $$f\in \frak m$$. Then we have $$\dim(A/fA)\geq \dim(A)-1$$. Moreover, equality holds if $$f$$ is not contained in any minimal prime ideal of $$A$$.

Theorem 2.5.23(a) Let $$A$$ be a finitely generated integral domain over $$k$$. Let $$p$$ be a prime ideal of $$A$$, we have $$\operatorname{ht}(p) + \dim(A/p) = \dim(A)$$.

1. $$A\to \mathcal{O}_{X,x}$$ is a local homomorphism of local rings, so it sends $$\mathfrak{m}_A$$ to $$\mathfrak{m}_x$$. $$t$$ is in the first by assumption, so it's in the second, and the maximal ideal of a local ring is precisely the non-invertible elements.
2. Since $$Y'\to Y$$ hits $$y$$ by construction, there are maps from $$\operatorname{Spec} k(x)$$ to $$Y'$$ and $$X$$ which agree after composing with the map to $$Y$$. So there's a map $$\operatorname{Spec} k(x)\to X'$$ which after composing with $$X'\to X$$ agrees with the map $$\operatorname{Spec} k(x)\to X$$ which picks out $$x$$. As $$X'\to X$$ is a closed immerison, this means $$x\in X'$$.
• Thank you for your answer. For (2), why there is a morphism $Spec k(x)\rightarrow Y'$(I understand there is $Spec k(y)\rightarrow Y'$). For (3), could we use the same argument to generalize theorem 2.5.23(a) to $A$ not necessary domain? – Hydrogen Mar 21 at 0:25
• The fact that $x$ maps to $y$ means there's a map of residue fields $k(y)\to k(x)$, or a map of spectra $\operatorname{Spec} k(x)\to\operatorname{Spec} k(y)$. Compose with the map you know. Secondly, no, not without further adjustments - consider something like $R=k[x,y,z]/(xy,xz)$: this has dimension $2$, but the quotient by the prime ideal $(z)$ gives the ring $k[x,y]/(xy)$ which is of dimension one. Thinking about the geometry of that should be instructive. – KReiser Mar 21 at 0:32
• It seems that the statement holds when $Spec(A)$ is irreducible but not necessary reduced. – Hydrogen Mar 21 at 3:46
• 1. Yes, $\operatorname{Spec} A$ irreducible means $A/Nil(A)$ is a domain. 2. $x$ is a closed point in a scheme locally of finite type over a field, apply Zariski's lemma. – KReiser Mar 21 at 6:06